example 2 evaluate a variable expression substitute – 2 for x and 7.2 for y. add the opposite –...

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EXAMPLE 2 Evaluate a variable expression Substitute 2 for x and 7.2 for y. Add the opposite – 2. = 7.2 + 2 + 6.8 aluate the expression y x + 6.8 when x = – 2 and 7.2. y x + 6.8 = 7.2 – (2) + 6.8 = 16 Add.

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Page 1: EXAMPLE 2 Evaluate a variable expression Substitute – 2 for x and 7.2 for y. Add the opposite – 2. = 7.2 + 2 + 6.8 Evaluate the expression y – x + 6.8

EXAMPLE 2 Evaluate a variable expression

Substitute – 2 for x and 7.2 for y.

Add the opposite – 2.= 7.2 + 2 + 6.8

Evaluate the expression y – x + 6.8 when x = – 2 and y = 7.2.

y – x + 6.8 = 7.2 – (–2) + 6.8

= 16 Add.

Page 2: EXAMPLE 2 Evaluate a variable expression Substitute – 2 for x and 7.2 for y. Add the opposite – 2. = 7.2 + 2 + 6.8 Evaluate the expression y – x + 6.8

One of the most extreme temperature changes in United States history occurred in Fairfield, Montana, on December 24, 1924. At noon, the temperature was 63°F. By midnight, the temperature fell to – 21°F. What was the change in temperature?

SOLUTION

The change C in temperature is the difference of the temperature m at midnight and the temperature n at noon.

EXAMPLE 3 Evaluate Change

Temperature

Page 3: EXAMPLE 2 Evaluate a variable expression Substitute – 2 for x and 7.2 for y. Add the opposite – 2. = 7.2 + 2 + 6.8 Evaluate the expression y – x + 6.8

Substitute values.

Add the opposite of 63.

= – 21 – 63

Write equation.

Find the change in temperature.

Write a verbal model. Then write an equation.

STEP 1

STEP 2

= – 84 Add – 21 and – 63.

Evaluate ChangeEXAMPLE 3

C m n= –

C m – n=

= – 21 + (– 63)

Page 4: EXAMPLE 2 Evaluate a variable expression Substitute – 2 for x and 7.2 for y. Add the opposite – 2. = 7.2 + 2 + 6.8 Evaluate the expression y – x + 6.8

The change in temperature was – 84°F.

ANSWER

EXAMPLE 3 Evaluate Change

Page 5: EXAMPLE 2 Evaluate a variable expression Substitute – 2 for x and 7.2 for y. Add the opposite – 2. = 7.2 + 2 + 6.8 Evaluate the expression y – x + 6.8

GUIDED PRACTICE for Example 2 and 3

Evaluate the expression when x = – 3 and y = 5.2.

4. x – y + 8 = – 0.2

5. y – (x – 2) = 10.2

6. (y – 4) – x = 4.2

A new car is valued at $15,000. One year later, the car is valued at $12,300. What is the change in value of car?

7. Car values

– $2700

ANSWER